Electric charge of `1 muC, -1 muC` and `2 muC` are placed in air at the corners A, B and C respectively of an equilateral triangle ABC having length of each side 10 cm. The resultant force on the charge at C is

Options

A. 0.9 N
B. 1. 8N
C. 2.7 N
D. 3.6 N

Answer

To find the resultant force on the charge at point C due to the charges at points A and B, we can follow these steps:

### Step 1: Identify the Charges and Their Positions
- Charge at A (qA) = +1 µC = +1 × 10^-6 C
- Charge at B (qB) = -1 µC = -1 × 10^-6 C
- Charge at C (qC) = +2 µC = +2 × 10^-6 C
- The distance between each pair of charges (length of sides of the triangle) = 10 cm = 0.1 m.

### Step 2: Calculate the Force on Charge C due to Charge A
Using Coulomb's Law:
\[ F_{CA} = k \frac{|q_C \cdot q_A|}{r^2} \]
Where:
- \( k \) (Coulomb's constant) = \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( r \) = 0.1 m

Substituting the values:
\[ F_{CA} = 9 \times 10^9 \frac{(2 \times 10^{-6})(1 \times 10^{-6})}{(0.1)^2} \]
\[ F_{CA} = 9 \times 10^9 \frac{(2 \times 10^{-12})}{0.01} \]
\[ F_{CA} = 9 \times 10^9 \times 2 \times 10^{-10} \]
\[ F_{CA} = 1.8 \, \text{N} \]

### Step 3: Calculate the Force on Charge C due to Charge B
Again using Coulomb's Law:
\[ F_{CB} = k \frac{|q_C \cdot q_B|}{r^2} \]
Substituting the values:
\[ F_{CB} = 9 \times 10^9 \frac{(2 \times 10^{-6})(1 \times 10^{-6})}{(0.1)^2} \]
Since \( q_B \) is negative, this force will be attractive:
\[ F_{CB} = 9 \times 10^9 \frac{(2 \times 10^{-12})}{0.01} \]
\[ F_{CB} = 9 \times 10^9 \times 2 \times 10^{-10} \]
\[ F_{CB} = 1.8 \, \text{N} \]

### Step 4: Determine the Direction of Forces
- \( F_{CA} \) (due to charge A) is repulsive and directed away from A towards C.
- \( F_{CB} \) (due to charge B) is attractive and directed towards B.

### Step 5: Calculate the Resultant Force
The angle between \( F_{CA} \) and \( F_{CB} \) is 120 degrees (since the triangle is equilateral). We can use the law of cosines to find the resultant force \( F_R \):
\[ F_R^2 = F_{CA}^2 + F_{CB}^2 + 2 F_{CA} F_{CB} \cos(120^\circ) \]
Since \( \cos(120^\circ) = -0.5 \):
\[ F_R^2 = (1.8)^2 + (1.8)^2 + 2(1.8)(1.8)(-0.5) \]
\[ F_R^2 = 3.24 + 3.24 - 3.24 \]
\[ F_R^2 = 3.24 \]
\[ F_R = \sqrt{3.24} \]
\[ F_R \approx 1.8 \, \text{N} \]

### Final Result
The resultant force on the charge at C is approximately **1.8 N**.

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